Meridians and Convergency

Meridians are great semi-circles that join the geographic North and South Poles. Because they all connect at the same two points, they cannot remain parallel to each other — they must converge at the poles. Understanding exactly how much they converge, and at what rate, is fundamental to great circle navigation.

Figure 1 — Meridians diverge from the North Pole, are parallel at the Equator, then converge again toward the South Pole

Figure 2 — At the Equator, any two meridians are exactly parallel to each other — convergency is zero

The Three Key Facts About Meridian Convergence

• At the Equator: any two meridians are exactly parallel. No matter how far apart they are in longitude, they never incline toward each other at the Equator. Convergency = 0°.
• At either Pole: meridians converge to a single point. The angle between two meridians at the Pole equals exactly their difference in longitude. If two meridians are 60° of longitude apart, they meet at the Pole at an angle of exactly 60°. Convergency = Change of Longitude.
• At intermediate latitudes: convergency is between 0° and the change of longitude — proportional to the sine of the latitude.

Figure 3 — At the Pole, the angle between two meridians equals exactly their change of longitude

Figure 4 — At intermediate latitude, convergency is proportionally between 0° (Equator) and ch.long (Pole)

The Convergency Formula

From the observations above, convergency varies between zero (at the Equator) and the full change of longitude (at the Poles). The factor that controls this is the sine of the latitude. This leads directly to the formula:

Why Sine of Latitude?

The sine function has exactly the right behaviour: sin(0°) = 0 at the Equator, and sin(90°) = 1.0 at the Poles. At 30°N, sin(30°) = 0.5, so convergency is exactly half the change of longitude. At 45°N, sin(45°) = 0.707, so convergency is about 70.7% of the change of longitude.

Figure 5 — Convergency shown as the angle of inclination between two meridians — with worked table showing values at Equator (0) and Pole (1.0)

The 30°N Worked Example

Consider two meridians 40° of longitude apart, at latitude 30°N:

Figure 6 — Tangents drawn to two meridians at 30°N — the angle between them is 20° (= 40° × 0.5)

When the Two Points are at Different Latitudes

In practice, most great circle routes go from one latitude to another — the two endpoints are not on the same parallel. In this case there is no single "latitude" to insert in the formula. The rigorous solution requires the mean latitude of the great circle route itself, which is closer to the nearer pole than the mid-latitude between the two endpoints.

However, the difference between sine of mid-latitude and sine of mean latitude is small unless the change of longitude is very large. For exam purposes and practical navigation, mid-latitude is used in the formula.

Great Circle Track and Convergency

At any point on the Earth, True North is defined by the direction of the local meridian. If you fly a great circle route, you will cross different meridians — and because those meridians are inclined to each other, your great circle track angle (measured from True North) will change continuously throughout the flight.

The amount by which a great circle track changes direction between two meridians is exactly equal to the convergency between those two meridians.

Figure 7 — A great circle cutting meridians X and Y — the angle 'a' at X and angle 'b' at Y differ by exactly the convergency between the meridians

Why a Rhumb Line Curves

A rhumb line (constant true track) appears as a straight line on a Mercator chart but as a curved line on a conformal conic chart. The reason is that to maintain a constant bearing from True North — and True North itself is rotating (because the meridians are converging) — the path must continuously curve. A rhumb line must compensate for meridian convergence in order to keep its track angle constant.

Conversely, a great circle — which is the shortest path between two points on the sphere — appears curved on a Mercator chart (curving toward the nearer pole) but as a straight line on a Lambert conformal chart (because the meridians themselves converge on that projection).

The D-I-I-D Rule

When flying a great circle route, the track angle changes continuously. The D-I-I-D diagram tells you the direction of that change for any combination of hemisphere and direction of travel.

The mnemonic D-I-I-D stands for the four cells read clockwise from top-left: Decreases — Increases — Increases — Decreases (starting with SH going West = Increases, going around). More usefully, remember the two rules for the Northern Hemisphere (where most of our traffic is):

How to Apply the Rule

Step 1: Calculate convergency (ch.long × sin mid-lat).
Step 2: Look up D-I-I-D for your hemisphere and primary direction of travel.
Step 3: Add or subtract convergency from the departure track to get the arrival track.
Step 4: Take the reciprocal of the arrival track to get the initial track from B to A.

Alternative: Parallel Construction Line

Draw the initial track from A. Parallel it across to B (draw a line at the same angle to the vertical). The convergency angle is then visible as the angle between this parallel line and the meridian at B. Add the two angles to get the track at B; take the reciprocal for the reverse direction. Both methods give the same answer.

Worked Examples 1–3

Worked Example 1 — Initial GC Track Between Two Points

Given: Initial GC track from A (40°00′N 002°00′W) to B (50°00′N 010°00′E) is 060°(T). Find the initial GC track from B to A.

Figure 8 — Diagram for Example 1 — A is on the left (W) meridian, B on the right (E) meridian; initial track 060° from A toward NE

Figure 9 — Checking the diagram: B is north and east of A, confirmed by latitude/longitude comparison

Step 1 — Draw the diagram. In the Northern Hemisphere, meridians converge northward. The initial track is 060°(T) — going NE. A is on the W meridian, B on the E meridian.

Step 2 — Check the diagram. B (50°N, 010°E) should be north and east of A (40°N, 002°W). ✓ North because lat B > lat A. ✓ East because 010°E is east of 002°W.

Step 3 — Calculate convergency.
Change of longitude = 002°W to 010°E = 12°
Mid-latitude = (40° + 50°) ÷ 2 = 45°N; sin 45° = 0.7071
Convergency = 12° × 0.7071 = 8.5°

Step 4 — Apply D-I-I-D. Northern Hemisphere, going East → track INCREASES. Track at A = 060°. Track at B (arrival) = 060° + 8.5° = 068.5°.

Step 5 — Reciprocal. Track from B to A = 068.5° + 180° = 248.5°(T)

Answer: Initial GC track from B to A = 248.5°(T)

Worked Example 2 — Finding an Unknown Longitude

Given: Initial GC track from C (36°00′N 015°00′E) to D (42°00′N) is 300°(T); final track at D is 295°(T). Find: (a) longitude of D; (b) GC track at 011°E.

Step 1 — Find convergency. Track changed from 300° to 295° = change of 5°. Going West in NH → track DECREASES ✓ (300° → 295°). Convergency = 5°.

Step 2 — Find change of longitude.
Mid-latitude = (36° + 42°) ÷ 2 = 39°N; sin 39° = 0.629
Convergency = ch.long × sin(39°)
5° = ch.long × 0.629 → ch.long = 5° ÷ 0.629 = 8°
C is at 015°E and going west → D is at 015° − 8° = 007°E

Figure 10 — Southern hemisphere meridians — meridians converge southward; for Q3 the setup reverses

Figure 11 — Parallel construction: extend track from H, add convergency angle at G to find reciprocal

Step 3 — Track at mid-longitude. Longitude 011°E is midway between 015°E and 007°E. Therefore the track is midway between 300° and 295° = 297.5°(T)

Answers: (a) Longitude of D = 007°E  ·  (b) Track at 011°E = 297.5°(T)

Conversion Angle

Consider the situation in the diagram below. Two meridians are 40° of longitude apart. A rhumb line connects A (30°N 020°W) to B (30°N 020°E) along the parallel of latitude — this RL has a constant direction of 090°(T) throughout.

A great circle connecting the same two points does NOT follow the parallel. Because the GC is the shortest path between the points on the sphere, it arcs toward the nearer pole (northward in the Northern Hemisphere for an easterly track). At A, the GC heads off on a direction less than 090°(T) — it starts by heading slightly north of east. At B, it arrives on a direction greater than 090°(T) — it arrives heading slightly south of east. At the mid-longitude (000°E/W), the GC direction = the RL direction = 090°(T).

Figure 12 — Rhumb line (090° constant) vs great circle — RL shown as straight, GC bows toward the north pole between A and B

Figure 13 — Same geometry on a different projection — GC appears straight, RL curves; relative positions unchanged

Deriving Conversion Angle

In the example above:

• Convergency = 40° × sin(30°) = 40° × 0.5 = 20°
• The GC track changed 20° between A and B
• The RL track is 090° throughout and equals the GC at the midpoint
• So at A, GC = 090° − 10° = 080°(T)
• At B, GC = 090° + 10° = 100°(T)
• The difference between GC and RL at either endpoint = 10°

This difference — between the great circle direction and the rhumb line direction at a given point — is called the Conversion Angle (CA). In this case, CA = 10°.

Notice that CA is exactly half the convergency. This is always the case (for routes between two points at the same latitude; it is also used as an approximation for routes at slightly different latitudes).

Which Direction to Apply CA?

The GC always bows toward the nearer pole. Therefore:

• In the Northern Hemisphere, the GC arcs northward → at the departure point, GC is more northerly (smaller track number for E-bound; larger for W-bound) than the RL.
• In the Southern Hemisphere, the GC arcs southward → at the departure point, GC is more southerly than the RL.

Equivalently, at the departure point: RL = GC ± CA. Use the D-I-I-D/logical reasoning to determine the sign for your specific situation.

Radio Bearings and Conversion Angle

Radio waves travel along great circle paths over the surface of the Earth. When a VOR, NDB, or VDF station measures a bearing to or from an aircraft, that bearing is a great circle direction. For navigation plotting on a Mercator chart (where rhumb lines appear as straight lines), you must convert from great circle to rhumb line before plotting.

The Correction

Apply the Conversion Angle to the GC bearing to obtain the RL bearing:

Important: Apply CA at the Measurement Point

This is a critical, frequently examined point. If the GC bearing is measured at the aircraft, apply CA at the aircraft's position. If measured at the station, apply CA at the station's position. The CA correction is position-specific.

Practical Example

If the GC track from A to B is 100°(T) and CA = 4°, the RL track at A is found by: GC is 100°, which is slightly south of east in the NH. The RL (which is more 'direct', less curved) will be slightly closer to 090° → RL = 100° − 4° = 096°(T)… but check the exact geometry for your specific question (see Worked Example 6 below).

Worked Examples 4–6

Worked Example 5 — INS Waypoint Changeover

Given: WP1 (53°N 030°W) → WP2 (53°N 020°W) → WP3 (53°N 010°W). An FMS/INS steers GC legs. What is the track change on passing WP2?

Figure 14 — INS waypoint changeover — WP1 to WP3 shows two consecutive GC legs at 53°N; a left turn occurs at WP2

Step 1 — RL track all the way. All three WPs are on 53°N. RL track = 090°(T) throughout.

Step 2 — GC legs. Each leg has ch.long = 10°, mean lat = 53°N, sin 53° = 0.799.
CA per leg = ½ × 10° × 0.799 = 4.0°
GC leg 1 (WP1→WP2): departs WP1 on 086°, arrives WP2 on 094°
GC leg 2 (WP2→WP3): departs WP2 on 086°, arrives WP3 on 094°

Step 3 — Track change at WP2. Approaching WP2 on 094°(T); at changeover, FMS reverts to 086°(T) for the new leg → change = 094° − 086° = 8° decrease. This is a left turn at the waypoint. Answer: 8° decrease (d)

Answer: (d) — an 8° decrease of track angle on passing WP2

Worked Example 6 — GC to RL Conversion at a Point

Given: A (55°N 000°E/W), B (54°N 010°E). Initial GC track from A to B is 100°(T). Find the RL track at A.

Figure 15 — Diagram for Example 6 — GC track 100°(T) at A; applying CA gives RL track 104°(T)

Step 1 — Calculate CA.
Ch.long = 10°; Mean latitude = (55° + 54°) ÷ 2 = 54.5°N; sin 54.5° = 0.814
CA = ½ × 10° × 0.814 = 4.07° ≈ 4°

Step 2 — Direction of correction. We're going SE (100°T) in the NH. The GC is MORE northerly than the RL (GC bows toward N pole). At A, GC < RL. Therefore RL = GC + CA = 100° + 4° = 104°(T)

⚠ Critical Point — Apply CA at the Measurement Position

The answer is 104°(T) because the GC bearing is measured AT A and we apply CA at A. If the GC bearing were given at B, we would apply CA at B's position instead. Never apply CA at the wrong endpoint.

Answer: (c) RL track at A = 104°(T)

Practice Questions

The following 10 questions are taken from the chapter. These are calculation-based questions — draw a diagram for each before calculating. Tap to reveal each answer.